The length of a wire is increased by `1 mm` on the application of a given load. If a wire of the same material, but of length and radius twice that of the first, on application of the same load, extension is

Options

A. 2 mm
B. 0.5 mm
C. 4 mm
D. 0.25 mm

Answer

B

Explanation

`Y=(F)/(pi r^(2)) xx (L)/(l_(1))`
`therefore l_(1)=(FL_(1))/(pi r_(1)^(2)y) and l_(2)=(FL_(2))/(pi r_(2)^(2)y)`
`(l_(2))/(l_(1))=(L_(2))/(L_(1)) xx (r_(1)^(2))/(r_(2)^(2))`